HDOJ 1003 Max Sum

最新推荐文章于 2021-03-06 02:42:32 发布

sunmoon1991

最新推荐文章于 2021-03-06 02:42:32 发布

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分类专栏： HDOJ 文章标签： output integer input each ini

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本文介绍了一个解决HDOJ1003最大子序列和问题的算法实现，通过输入一系列整数，计算并输出该序列中能够构成的最大连续子序列及其和，同时给出子序列的起始和结束位置。

HDOJ 1003 Max Sum

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

KEY:

#include <iostream>

using namespace std;

int main()
{
    int i,j,n,t,max,sum,startnum,endnum,num,temp;
    cin>>t;
    for(int k=1;k<=t;k++)
    {
        cin>>n;
        max=-999999;
        sum=0;
        temp=1;
        for(i=0;i<n;i++)
        {
            cin>>num;
            sum=sum+num;
            if(sum>max)
            {
                max=sum;
                startnum=temp;
                endnum=i+1;
            }
            if(sum<0)
            {
                sum=0;
                temp=i+2;
            }
        }
        cout<<"Case "<<k<<':'<<endl;
        cout<<max<<' '<<startnum<<' '<<endnum<<endl;
        if(k!=t)
        cout<<endl;
    }
    return 0;
}